Ž .JOURNAL OF ALGEBRA 188, 648]685 1997ARTICLE NO. JA966860

Ž .The Gelfand]Graev Representation of U 3, q

Julianne G. RainboltU

( )Department of Mathematics, Statistics, and Computer Science mrc 249 ,Uni ersity of Illinois at Chicago, 851 South Morgan Street, Chicago, Illinois 60607-7045

Communicated by George Glauberman

Received June 20, 1996

In this paper we explicitly calculate the irreducible representations of theendomorphism algebra of the Gelfand]Graev representation of the unitary groupŽ .U 3, q . In addition, we compute the structure constants of this endomorhphism

algebra. Q 1997 Academic Press

INTRODUCTION

Ž .Let GL 3, F denote the general linear group of invertible 3 by 3q

Ž . Ž .matrices over F . Let F : GL 3, F ª GL 3, F denote the homomorphismq q q

Ž . ŽŽ q . t.y1defined by F a s a . Then F is a Frobenius map and the groupi j i jŽ .of fixed points of F is the unitary group U 3, q . In this paper we will

examine the structure and representations of the Hecke algebra H of theŽ .Gelfand]Graev representation of U 3, q . In this case, the Gelfand]Graev

representation is the induced representation of any nontrivial linear repre-Ž . Ž .sentation of the maximal unipotent subgroup U of U 3, q to U 3, q . The

Ž .center of GL 3, F is connected, thus there is only one Gelfand]GraevqŽ . w xrepresentation G of U 3, q 3, p. 519 , and G is independent of the choice

of the nontrivial linear representation of U.After a discussion of some preliminary results in Section 1, Section 2

contains the calculations that explicitly give the irreducible representationsw xof H. D. Surowski 8 has previously calculated some of these irreducible

Ž . Ž .representations in the case of SU 3, q instead of U 3, q . The techniquesŽused in Section 2 are different than Surowski’s most notably the use of

.Curtis’s theorem to be discussed below , and this section provides a

* E-mail address: Rainbolt@math.uic.edu.

648

0021-8693r97 $25.00Copyright Q 1997 by Academic PressAll rights of reproduction in any form reserved.

GELFAND]GRAEV REPRESENTATION 649

complete list of the irreducible representations. The structure of theHecke algebra is further examined in Section 3 and the structure constantsof the algebra are explicitly calculated.

The material in this paper forms a part of the author’s doctoral disserta-tion at the University of Illinois at Chicago. I thank my advisor BhamaSrinivasan for her invaluable assistance. In addition, I thank JonathanAlperin, Paul Fong, David Radford, Steve Smith, and Jeremy Teitelbaumfor their comments and discussions on this material. I thank Charles Curtisfor suggesting this problem. I also thank Henri Gillet for financial supportin the form of a research assistantship.

1. PRELIMINARY RESULTS ANDBACKGROUND INFORMATION

1.1. Gelfand]Grae Representations

˜Let G be a connected reductive algebraic group defined over a finite˜Ffield F . Given a Frobenius endomorphism F, let G s G , the fixed pointsq

of F. The Gelfand]Graev characters of G are constructed in the followingŽ w x.way see, for example 5, Chap. 7; 3, pp. 518]519 . Given an F-stable

˜ ˜ ˜ ˜Borel subgroup B of G and an F-stable maximal torus T contained in B,0˜ ˜we have the root system F of G with respect to T . Let P denote the set0

˜ ˜of simple roots in F corresponding to B. Let U be the unipotent radical of˜ ˆ ˜B and let U be the subgroup of U generated by the root subgroups

� 4corresponding to the nonsimple roots of F. Also let U be the set ofa a g P

simple root subgroups. Then for each F-orbit i on P let

U s U .Łi aagi

˜F ˆFLet c be a linear character of U s U which is trivial on U . Then c iscalled nondegenerate if c is nontrivial when it is restricted to the F-fixedpoints of any of the U . Given a nondegenerate character c of U, leti

GŽ .G s Ind c . Then G is called a Gelfand]Graev character of G.U

1.2. Properties of Gelfand]Grae Representations

w xThe Gelfand]Graev characters G are multiplicity free 7, Theorem 49 .Thus, in principle, it might be feasible to decompose the Gelfand]Graev

˜Ž .representation into irreducible representations. In the case that Z G iswconnected there is only one distinct Gelfand]Graev representation 3, p.

x519 . Thus, in this case the choice of nondegenerate linear representation˜of U will not affect the results. Let T denote a maximally split F-stable0

JULIANNE G. RAINBOLT650

˜maximal torus of G. A maximal torus of G is defined to be a subgroup of˜F ˜ ˜the form T where T is an F-stable maximal torus of G. Now the

˜G-conjugacy classes of F-stable maximal tori of G are parametrized by the˜ ˜Ž .F-conjugacy classes of N T rT where the F-conjugate of x by g isG 0 0

Ž .y1 Ž w x.defined to be gxF g see, for example, 1, Propositions 3.3.2 and 3.3.3 .˜w x Ž .Also, given an F-conjugacy class x , T is conjugate in G to a maximalx

torus of G where

˜ y1T s A g T N xAx s F A .Ž .� 4x 0

˜Ž .Conversely, all the maximal tori of G are conjugate in G to T for somexx.

˜ ˜ GGiven an F-stable maximal torus T of G let R denote theT , u

Deligne]Lusztig generalized character, where u is an irreducible character˜F GŽ w x.of the torus T s T see, for example, 1, Chap. 7 . Let Q denote theT

Green function, which is defined for all unipotent elements u g G byGŽ . G Ž . Ž w x.Q u s R u see, for example, 1, p. 212 .T T , u

˜Ž .Given a pair T , u there exists a unique irreducible character x of GT , u

² : ² G :such that x , G / 0 and x , R / 0. Also any irreducible char-T , u T , u T , u

² :acter x of G such that x , G / 0 coincides with a x for some pairT , u

˜ ˜Ž . Ž w x. Ž .T , u see, for example, 3, Theorem 2.1 . The pairs T , u are partitionedŽ w x.into geometric conjugacy classes see, for example, 1, Sect. 4.1 . In fact,

each Gelfand]Graev character of G is equal to

e e RG˜ ˜G T T , uÝ G GR , RŽ .T , u T , u˜Ž .T , u gk

mod G

˜Ž .for some geometric conjugacy class k of pairs T , u where e sG˜rel. rank GŽ . Ž w x.y1 see, for example, 1, Proposition 8.4.7 . In particular when x² : ² G :is cuspidal, x , G s x , R s 1.T , u

1.3. The Hecke Algebra

˜FGiven G s G , the maximal unipotent subgroup U of G and a nonde-generate linear character c of U, the Hecke algebra H is constructed in

Ž w x .the following way. This can be done more generally, see 4, Sect. 11 . Lete denote the central primitive idempotent in CU corresponding to c . Thatis,

< <y1 y1e s U c u u.Ž .ÝugU

Then H s eCGe.

GELFAND]GRAEV REPRESENTATION 651

Let N be a set of double coset representatives of U, U in G. LetŽ . < < < < < n < � nind n s UnU r U s U : U l U for n g N. Let J s n g N N c s c on

n 4 Ž . � 4U l U . Let c s ind n ene. Then c is a basis for H called then n ng Jw xstandard basis of H 4, Proposition 11.30 . There is a bijection from the set

² :of irreducible characters x of G such that x , G / 0 to the set of allirreducible characters of H. This bijection is given by restriction from CG

w xto H 3, Proposition 2.2 . Also the primitive central idempotents of H are� 4ee where e is a primitive central idempotent of CG associated with a x

² :such that x , G / 0. Since G is multiplicity free, the Hecke algebra H isw xcommutative 3, Proposition 2.2 . Thus these idempotents are actually

Ž .primitive non-central idempotents. Thus they give us the simple modulew xCGee which affords x 4, Corollary 11.27 .

1.4. Curtis’s Theorem

As mentioned above, the set of irreducible characters of H is in² :bijection with the set of irreducible characters x of G such that x , G /

˜Ž .0. Namely, given a pair T , u there exists a unique irreducible character² : ² G :x of G such that x , G / 0 and x , R / 0. The restrictionT , u T , u T , u T , u

of x to H is the corresponding irreducible character f of H. ThusT , u T , u˜Ž .we can index the characters of H by the pairs T , u . In addition,

˜ X XŽ . Ž .X Xf s f if and only if the pairs T , u and T , u are geometricallyT , u T , u

w xconjugate 4, Theorem 3.1 . So each irreducible representation f of HT , u

corresponds to the unique irreducible character x of G which occurs asT , u

a common constituent of both RG and G.T , u

w xCurtis’s theorem is as follows 3, Theorem 4.2 :

˜Ž .THEOREM 1.1. Let the pair T , u , the Gelfand]Grae representationGŽ .G s Ind c , and the Hecke algebra H be gi en. Let u denote the extensionU

of u to CT. Also let x and x denote the semisimple and unipotent parts ofs ux g G. Then:

Ž .i There exists a unique homomorphism f : H ª CT , independent ofTu , which has the property that each character f : H ª C can be factored asT , u

f s u ? f .T , u T

Ž . Ž . Ž .Ž .ii f c s Ý f c t t where c is an element in the standardT n t g T T n nŽ .Ž .basis of H described abo¨e and the coefficients f c t are gi en byT n

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gung .Ž . Ž . Ž . Ž .Ž .ÝT n T uG² : < < < <Q , G U C tŽ .T G ggG , ugUy1Ž .gung sts

1.2Ž .

JULIANNE G. RAINBOLT652

Note that Curtis’s theorem says that we have the following commutativediagram:

fT 6

H CT

6

fT , uu6

C

Thus if we first find the homomorphisms f for each maximal torus T ofTG and then compose these with the irreducible characters of T we will getall the irreducible characters of H.

2. THE GELFAND]GRAEV REPRESENTATION OF G

2.1. Notation

Ž . Ž .As described in the Introduction let F : GL 3, F ª GL 3, F denoteq qŽ . ŽŽ q . t.y1the twisted Frobenius map, defined by F a s a . From now oni j i j

F˜ ˜Ž .GL 3, F will be denoted by G. Instead of using the fixed point group Gq˜Fit will be convenient to take the unitary group G given by conjugating G

by

0 0 1.0 y1 0ž /1 0 0

˜ ˜Now let U denote the unipotent subgroup of G which consists of upperunitriangular matrices. That is,

¡ ¦1 t u~ ¥U s t , u , ¨ g F .0 1 ¨ q¢ §ž /0 0 1

˜FLet U be the subgroup of G given by conjugating U by

0 0 1.0 y1 0ž /1 0 0

Then

¡ ¦1 t uqq1 qq~ ¥2U s t , u g F and t s u q u .0 1 t q¢ §ž /0 0 1

GELFAND]GRAEV REPRESENTATION 653

w x w xLet U, U denote the commutator subgroup of U. Since Ur U, U isisomorphic to the additive group F 2 , there is a correspondence betweenqthe irreducible linear characters c of U and the additive characters x ofF 2 . This correspondence is given byq

1 t uqc s x t .Ž .0 1 tž /0 0 1

Choose any nontrivial irreducible linear character c of U. Let G sGŽ .Ind c . Then G is independent of the choice of c and G is theU

Ž w x.Gelfand]Graev character of G see 3; 5, Chap. 14 . Let e denote theidempotent

1y1e s c u u.Ž .Ý< <U ugU

Ž .Then let H be the Hecke algebra e CG e.From now on, the diagonal matrix

a 0 00 b 0ž /0 0 c

w xwill be denoted by a, b, c . In addition, the element

0 0 a0 b 0ž /c 0 0

$w xwill be denoted by a, b , c . Also an element

1 t uq g U0 1 tž /0 0 1

w xwill be denoted by t, u .The following is a summary of the above notation:

G s GL 3, Fž /q

$ $F˜G s U 3, q s 1, y1, 1 G 1, y1, 1Ž .

w x qq1 q2U s t , u N u , t g F and t s u q u� 4q

c s nontrivial linear character of U

JULIANNE G. RAINBOLT654

w x2x s additive character of F such that c t , u s x tŽ .Ž .q

G s IndG cŽ .U

< < y1e s 1r U c u uŽ .Ž . ÝugU

H s e CG e.Ž .

We recall Curtis’s theorem, Theorem 1.1. The main result of this sectionwill be the computations of the unique homomorphisms f : H ª CT , forT ii

each maximal torus T of G.i

2.2. The Maximal Tori of G

˜ ˜Let T denote the maximally split F-stable maximal torus of G:

U˜ w xT s t , u , ¨ N t , u , ¨ g F .½ 5q

As discussed in the Introduction, the G-conjugacy classes of F-stable˜ ˜ ˜Ž .maximal tori of G are parametrized by the F-conjugacy classes of N T rTG

˜ ˜ ˜Ž w x. Ž . Ž .see, for example, 5, Proposition 3.23 . Also N T rT s W T , the WeylGgroup, which in this case is isomorphic to S , the symmetric group on three3

Ž .elements. Let, e.g., 12 denote the element of S corresponding to the3permutation matrix

0 1 0,1 0 0ž /0 0 1

Ž .which will also be denoted by 12 , and similarly for the other permutationmatrices.

²Ž . Ž .:Now since S s 12 , 23 , the twisted Frobenius map F can be thought3Ž . Ž . Ž . Ž .of as the automorphism of S that takes 12 to 23 and takes 23 to 12 .3

w xLet x denote the F-conjugacy class of x. Then

w xe s e, 123 , 132� 4Ž . Ž .12 s 12 , 23� 4Ž . Ž . Ž .13 s 13 .� 4Ž . Ž .

Thus there are three classes of maximal tori in G. Each of these maximal˜Ž . w xtori is conjugate in G to T for some F-conjugacy class x wherex

˜ y1T s A g T N xAx s F A .Ž .� 4x

GELFAND]GRAEV REPRESENTATION 655

Thus the three maximal tori are

w x w x w yq yq yq xT s t , u , ¨ ¬ t , u , ¨ s ¨ , u , t and t , u , ¨ / 0 ,� 4e

w x w x w yq yq yq xT s t , u , ¨ ¬ u , t , ¨ s ¨ , u , t and t , u , ¨ / 0 ,� 4Ž12.

and

w x w x w yq yq yq xT s t , u , ¨ ¬ ¨ , u , t s ¨ , u , t and t , u , ¨ / 0 .� 4Ž13.

From now on denote T by T , T by T , and T by T .e 0 Ž12. 1 Ž13. 2The above conditions on T , T , and T simplify to0 1 2

w yq x q 2y1 qq1T s t , u , t N t s 1, u s 1� 40

w 2 yq x q 3q1T s t , t , t N t s 1� 41

w x qq1 qq1 qq1T s t , u , ¨ N t s 1, u s 1, and ¨ s 1 .� 42

ˆNotice that T is not a subgroup of G but there exists a subgroup T of1 1ˆ ˜G such that T is conjugate to T in G. Since T is diagonal, it will be1 1 1

ˆ < <convenient to work with the group T instead of T . Notice that T s1 1 0Ž .Ž 2 . < < 3 < < Ž .3q q 1 q y 1 , T s q q 1 and T s q q 1 . Also notice that T is1 2 0the maximally split torus in G.

2.3. The Hecke Algebra$

w x Ž . � 4Let s s 1, y1, 1 . Then N T s A N A g T or A g T s , whereG 0 0 0Ž .N T denotes the normalizer of T in G. We have the Bruhat decompo-G 0 0

sition

G s UT wU.D 0ws1, s

But then a basis of the Hecke algebra is given by

c s ind x ex e N x g T j T s andŽ .½ j j j j 0 0

c y s c x yxy1 for all y g U l x Uxy1Ž . 5Ž .j j j j

Ž w x. Ž . < x <see, for example, 4, Proposition 11.30 . Here ind x s U : U l U s thenumber of left cosets of U in the double coset UxU. The following arecalculations that determine this basis explicitly.

w x qq1 q2Fix a y g U, i.e., y s ¨ , w for some ¨ , w g F and ¨ s w q w .q$ 2yq q y1w xFirst let x g T s. That is, x s t, yu, t for some t, u such that t s 10

JULIANNE G. RAINBOLT656

and uqq1 s 1. Then

1 0 0y1 y1 qxyx s .yut ¨ 1 0� 0yqy1 y1 yqwt yu ¨t 1

Thus xyxy1 g U if and only if uty1 ¨ q s 0, wtyqy1 s 0, and uy1 ¨tyq s 0.But t, u / 0 so that ¨ s 0 and w s 0. Thus xU l U s I when x g T s.0

Ž . < < 3Thus ind x s U s q when x g T s.0w yq x y1Now let x g T . So x s t, u, t and xUx s U. But note that the0

Ž . Ž y1 . y1condition that c y s c xyx for all y g U l xUx implies that

w x w y1 qq1 xc ¨ , w s c tu ¨ , t w .Ž . Ž .But this holds if and only if tuy1 ¨ s ¨ for all ¨ g F 2 . This holds if andq

only if t s u which causes tyq s uyq s u since uqq1 s 1. Thus we musthave

w x qq1x g t , t , t ¬ t s 1 .� 4y1 x Ž .But then xyx s y. Also for x of this type U l U s U and ind x s 1.

Thus if$ $ $

yq yq 3 yqc s ind t , yu, t e t , yu, t e s q e t , yu, t ež /u , t

andw x w x w xc s ind u , u , u e u , u , u e s u , u , u e,Ž .u

then the set

c , c N t , u g F 2U and uqq1 s 1� 4u , t u q

is a basis for the Hecke algebra H. Note that the number of elements inŽ . Ž 2 .Ž . 2Ž .this basis is q q 1 q q y 1 q q 1 s q q q 1 .

Ž . ² G :2.4. Calculations of C t , t g T , and Q , GG i Ti

Ž .In order to apply Curtis’s theorem we need to know C t for t in aG² G :maximal torus T of G. Also we will need the values of Q , G for each ofi Ti

the maximal tori.For all three tori if t g T and the entries on the main diagonal are alli

Ž .distinct then C t s T . Also, if the entries on the main diagonal are allG iŽ .the same, then clearly C t s G.G

w yq xFix a t g T so that t s t , t , t for some fixed t , t such that0 1 2 1 1 2q 2y1 qq1 yq Ž yq .t s 1 and t s 1. Suppose that t s t and t / t so t / t .1 2 1 1 1 2 1 2

Thena 0 c¡ ¦~ ¥0 e 0C t s .Ž .G ¢ §� 0g 0 i

GELFAND]GRAEV REPRESENTATION 657

Let

a 0 c¡ ¦~ ¥0 1 0H s .2 ¢ §� 0g 0 i

Ž .Then H ( U 2, q under the isomorphism2

a 0 c¡ ¦ a c~ ¥0 1 0b : ª .½ 5ž /g i¢ §� 0g 0 i

Also let

w x qq1S s 1, w , 1 N w s 1 .� 42

Then S ( S where S is the subgroup of F 2U consisting of elements x such2 q

qq1 Ž . Ž .that x s 1. Thus, in this case, C t s H = S ( U 2, q = S. SoG 2 2< Ž . < < Ž . < < < Ž .Ž 2 .Ž . Ž .3Ž .C t s U 2, q S s q q q 1 q y 1 q q 1 s q q q 1 q y 1 .G

yq Ž yq .Note that the case t s t and t / t so t / t cannot occur. Also1 2 1 1 1 2yq Ž yq .the case t s t and t / t so t / t cannot occur.2 1 1 2 1 1

w q 2 yq xNow let t g T so that t s t , t , t for some fixed t such that1 1 1 1 1t q 3q1 s 1. Note that it is not possible that exactly two of the entries on the1main diagonal of t are the same.

w xNow fix a t g T so that t s t , t , t for some fixed t , t , t such that2 1 2 3 1 2 3qq1 Ž .t s 1. Suppose that t s t and t / t so t / t . Theni 1 2 2 3 1 3

a b 0C t s .Ž . d e 0G ½ 5ž /0 0 i

Let

a b 0H s .d e 01 ½ 5ž /0 0 1

Ž .Then H ( U 2, q under the isomorphism1

a b 0 a ba : ª .d e 0 ž /d ež /0 0 1

Also let S be the torus1

w x qq11, 1, w N w s 1 .� 4

JULIANNE G. RAINBOLT658

Then S is isomorphic to the multiplicative subgroup S of F 2U consisting of1 q

qq1 Ž .elements x such that x s 1. So in this case C t s H = S (G 1 1Ž .U 2, q = S.

Ž .Now suppose that t s t and t / t so t / t . Then1 3 2 3 1 2

a 0 c¡ ¦~ ¥0 e 0C t s .Ž .G ¢ §� 0g 0 i

Ž . Ž .So again C t s H = S ( U 2, q = S.G 2 2Ž .Now suppose that t s t and t / t so t / t . Then2 3 1 2 1 3

a 0 0¡ ¦~ ¥0 e fC t s .Ž .G ¢ §� 00 h i

So now let

1 0 0¡ ¦~ ¥0 e fH s3 ¢ §� 00 h i

and

w x qq1S s w , 1, 1 ¬ w s 1 .� 43

Ž . Ž .Then in this case C t s H = S ( U 2, q = S.G 3 3Notice that when t g T has exactly two entries on the diagonal the2

< Ž . < < Ž . < < < Ž .Ž 2 .Ž . Ž .3same, C t s U 2, q S s q q q 1 q y 1 q q 1 s q q q 1GŽ .q y 1 .

LEMMA 2.1. We ha¨e that

² G :Q , G s 1T0

² G :Q , G s y1T1

² G :Q , G s y1.T2

GŽ .Proof. For any maximal torus T we have, by definition, Q u sTG Ž . GR u for u g U. But also, when u is in general position, R s "x forT , u T , u

Ž G Ž .some irreducible character x of G. The sign is chosen so that R u s 1T , u

.when u is in the regular unipotent conjugacy class. Thus if we choose the² G : ² G :character u of T appropriately we get that Q , G s R , G s "1.i i T T , ui i i

The correct sign can be determined by examining the character table for Gw xin, for example, 6 .

GELFAND]GRAEV REPRESENTATION 659

2.5. Characteristic Equations of Elements of T0

Recall, as mentioned in the Introduction, given an F-stable maximal˜ ˜Ftorus T and a character u of T s T there exists a unique homomorphism

f : H ª CT , independent of u , which has the property that each homo-TŽ .morphism f : H ª C can be factored as f s u f . In addition, f cT , u T , u T T n

Ž .Ž .s Ý f c t t, where c is an element in the basis of H given above,t g T T n nand

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gungŽ . Ž . Ž . Ž .Ž .ÝT n T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

w x3, Theorem 4.2 . Here n is the element in Z j T w such that c s0 n$qq1Ž . �w x 4 w xind n ene, Z s u, u, u N u s 1 , and w s 1, y1, 1 .

In order to explicitly compute these homomorphisms, it remains toŽ y1 .determine for which g g G and u g U we have that gung s t, for as

fixed t g T.Note. Since t is semisimple, t and un have the same characteristic

Ž y1 .equation if and only if there exists a g such that gung s t. So it issenough to find conditions on u and n so that un and t have the samecharacteristic equation. In this section we will determine under whatconditions un and t have the same characteristic equation when t g T .0

The following lemma is a general result about all three maximal tori ofG.

LEMMA 2.2. Let t be an element of a maximal torus T . Also let u g Uiw x � Ž y1 . 4and let n s n s l, l, l . Then u g U N gun g s t s U if all thel l s

� Ž y1 . 4diagonal entries of t are equal to l, otherwise u g U N gun g s t s B.l s

Proof. It is clear that un and t have the same characteristic equation ifŽ y1 . Ž y1 y1. y1and only if t s n. For this t, gung s gug gng s gng s n s ts s

for all u g U and for all g g G.

˜In what follows let D denote the set of diagonal matrices in G.

Uw xLEMMA 2.3. Fix an element z in D, so z s a, b, c for some a, b, c g F .qw xAgain let u be an element of U, so u s r, s for some r and s such that

r qq1 s s q sq. But now let n g T w, so that0

$yqn s n s m , yl, m ,l, n

JULIANNE G. RAINBOLT660

for some l, m such that lqq1 s 1. Then un and z ha¨e the same characteris-tic equation if and only if the following three conditions occur:

s s a q b q c q l mq 2.4Ž . Ž .r qq1 s ab q bc q ac mq q m q sl ly1 2.5Ž . Ž .Ž .mqy1 s lay1 by1cy1 . 2.6Ž .

Proof. Note that un and z will have the same characteristic equation ifand only if

det xI y un s x y a x y b x y cŽ . Ž . Ž . Ž .s x 3 y a q b q c x 2 q ab q bc q ac x y abc.Ž . Ž .

Now

det xI y un s x 3 y smyq y l x 2Ž . Ž .y myqq1 q smyql y r qq1myql x y lmyqq1.Ž .

Equating coefficients gives the three equations

smyq y l s a q b q c 2.7Ž .myq m q sl y r qq1l s y ab q bc q ac 2.8Ž . Ž .Ž .

lmyqq1 s abc. 2.9Ž .

Ž . Ž .These are the same equations as 2.4 ] 2.6 above.

w yq xCOROLLARY 2.10. Fix an element t in T , so t s t , t , t for some0 1 2 1t , t g F 2

U with t qq1 s 1. Using the same notation as in the preceding lemma,1 2 q 2

un and t ha¨e the same characteristic equation if and only if the following twoconditions occur:

s s t q tyq q t q l mq 2.11Ž .Ž .1 1 2

mqy1 s lt qy1 ty1 . 2.12Ž .1 2

yq Ž .Proof. Substituting t s a, t s b, and t s c immediately gives 2.111 2 1Ž . Ž . Ž . Ž . Ž .and 2.12 from 2.4 and 2.6 of the preceding lemma. That is, 2.4 ] 2.6

of the above lemma become

s s t q t q tyq q l mq 2.13Ž .Ž .1 2 1

r qq1 s t t q t tyq q tyqq1 mq q m q sl ly1 2.14Ž .Ž .Ž .1 2 2 1 1

mqy1 s lt qy1 ty1 . 2.15Ž .1 2

GELFAND]GRAEV REPRESENTATION 661

Ž . Ž . Ž .Thus it remains to show that 2.14 is implied by 2.13 and 2.15 . NowŽ . qq1 q y1Ž yq yqq1. y12.14 requires that r s m l t t q t t q t q ml q s. By1 2 2 1 1the defining conditions on U we already have the requirement r qq1 s s q

q Ž .s . Thus, from 2.13 ,

r qq1 s s q sq s s q t q q t q q tyq 2 q lq mq 2Ž .1 2 1

s s q t q q ty1 q ty1 q ly1 m ,Ž .1 2 1

q 2 U qq1 qq1 Ž .2since x s x for all x g F and since t s l s 1. Thus by 2.15 ,q 2

r qq1 s mqly1 t t q mqly1 t tyq q mqly1 tyqq1 q mly1 q s1 2 2 1 1

Ž . Ž . Ž .as required. Thus 2.13 and 2.15 imply 2.14 .

The following lemma and proof are a modification of the lemma andw xproof in 2, p. 500 .

LEMMA 2.16. Let z, u and n be as in Lemma 2.3. Suppose that unl, m l, m

and z ha¨e the same characteristic equation. Then r s 0 if and only if yl isan eigen¨alue of z.

Ž . qq1Proof. Using 2.5 of Lemma 2.3, we have that r s 0 if and only ifŽŽ . q . y1 Ž . Ž .ab q ac q bc m q m q sl l s 0. Thus, using 2.4 and 2.6 , we have

qq1 Ž . 2Žthat r s 0 if and only if 0 s y ab q ac q bc l y abc y l a q b q c. 3 Ž . 2 Žq l . But this is true if and only if 0 s yl y a q b q c l y ab q ac.q bc l y abc. But as noted above the characteristic equation of z is

3 Ž . 2 Ž .x y a q b q c x q ab q ac q bc x y abc. Thus yl is a root of thecharacteristic equation for z if and only if r s 0.

COROLLARY 2.17. Let t be an element of a maximal torus T . Also letiu g U, and n g T w. Also suppose that un and t ha¨e the samel, m 0 l, m

characteristic equation. Then r s 0 if and only if yl is an eigen¨alue of t.

Proof. This is immediate from Lemma 2.16.

In the remainder of this section we will distinguish the results betweenŽwhen t g T has a characteristic equation with three distinct roots t /0 1

yq yq. Ž yq .t , t / t , t / t , two equal roots t s t but t / t , and three2 2 1 1 1 1 1 2 1Ž yq . Ž yq .equal roots t s t s t . These cases will be denoted by t , t , t ,1 1 2 1 2 1

Ž . Ž . Ž yqt , t , t , and t , t , t , respectively. Note that the cases t s t , t / t1 2 1 1 1 1 2 1 1 1yq .and t s t , t / t are not possible, as mentioned in Subsection 2.4.2 1 2 1

Ž . Ž .Note that there are q q 1 elements t in T in case t , t , t , q q q 10 1 1 1Ž . Ž 2 .Ž . Ž . Želements t in case t , t , t , and thus q y 1 q q 1 y q q 1 y q q q1 2 1

. Ž .2Ž . Ž yq .1 s q q 1 q y 2 elements t in case t , t , t .1 2 1Ž .Note that the notation a, b, c means that a, b, and c are distinct.

JULIANNE G. RAINBOLT662

w yq xPROPOSITION 2.18. Let t s t , t , t g T be fixed. Then the number1 2 1 0of u g U and n g T w such that un and t ha¨e the same characteristic0equation is the following:

Ž . Ž yq . ŽŽ . Ž .Ž ..1 If t is of type t , t , t : q y 1 q q q y 1 q q 1 .1 2 1

Ž . Ž . Ž Ž . Ž .2Ž ..2 If t is of type t , t , t : 2 q y 1 q q y 1 q q 1 .1 2 1

Ž . Ž . ŽŽ . Ž .Ž ..3 If t is of type t , t , t : q y 1 q q q y 1 q q 1 .1 1 1

Proof. As above, we use the notation n s n . Now for all three typesl, m

Ž .of t there are q q 1 choices for l and by 2.12 , there are q y 1 choicesŽ .for m. Also by 2.11 , s is completely determined by m, l, and t and by

Ž yq .Corollary 2.17 if yl is chosen to be an eigenvalue i.e., one of t , t , t1 2 1then r s 0. Also if yl is not an eigenvalue, then r qq1 s s q sq / 0. So ifyl is not an eigenvalue there are q q 1 choices for r. Thus suppose t is

Ž yq .of type t , t , t . Then if l is chosen so that yl is an eigenvalue, then1 2 1l s yt , since t / tyq. Thus when yl is an eigenvalue, there is 1 choice2 1 1for l, q y 1 choices for m, 1 choice for s, and 1 choice for r. Also if yl isnot an eigenvalue, there are q q 1 y 1 s q choices for l, q y 1 choices

Ž .for m, 1 choice for s, and q q 1 choices for r. This gives 1 . Very similarŽ . Ž . Ž .reasoning gives results 2 and 3 . The only differences are that in 2 , if

Ž .yl is an eigenvalue then it is either t or t and in 3 , if yl is an1 2eigenvalue then it is t .1

The above proposition is a modification of a similar proposition forŽ . w xGL 3, q proved by Carter in 2, Proposition 5 .

2.6. Characteristic Equations of Elements of T and T1 2

Recall that T denotes the maximal torus of elements of the form1w q 2 yq x q 3q1t s t , t , t such that t s 1. Thus there are two types of elements1 1 1 1

q 2 yq q 2 yq yq q 2 Žin T , either t / t , t / t , and t / t or t s t s t . The1 1 1 1 1 1 1 1 1 1q 3q1 .condition that t s 1 forces all three to be equal if any two are equal.1

Ž q 2 yq . Ž .These two types will be denoted by t , t , t and t , t , t , respectively.1 1 1 1 1 1Ž . 3Notice that there are q q 1 elements t of type t , t , t and q q 1 y1 1 1

Ž . Ž 2 . Ž q 2 yq .q q 1 s q q y 1 elements t of type t , t , t .1 1 1

w q 2 yq xPROPOSITION 2.19. Fix an element t in T , so t s t , t , t for some t1 1 1 1 1such that t q 3q1 s 1. Using the same notation for u and n as in the pre¨ious1 l, m

section, un and t ha¨e the same characteristic equation if and only if thel, m

following two conditions occur:

s s t q t q 2 q tyq q l mq 2.20Ž .Ž .1 1 1

mqy1 s ltyq 2qq y1. 2.21Ž .1

GELFAND]GRAEV REPRESENTATION 663

q 2 yq Ž .Proof. Substituting t s a, t s b, and t s c immediately gives 2.201 1 1Ž . Ž . Ž . Ž . Ž .and 2.21 from 2.4 and 2.6 of Lemma 2.3. That is, 2.4 ] 2.6 of Lemma

2.3 become

s s t q t q 2 q tyq q l mq 2.22Ž .Ž .1 1 1

r qq1 s t q 2q1 q t q 2yq q tyqq1 mq q m q sl ly1 2.23Ž .Ž .ž /1 1 1

mqy1 s ltyq 2qq y1. 2.24Ž .1

Ž . Ž . Ž .Thus it remains to show that 2.23 is implied by 2.22 and 2.24 . NowŽ .2.23 requires that

r qq1 s mqly1 t q 2q1 q t q 2yq q tyqq1 q mly1 q s.Ž .1 1 1

Ž .From 2.22 we have that

r qq1 s s q sq s s q t q q t q 3 q tyq 2 q lq mq 2Ž .1 1 1

s s q t q q ty1 q tyq 2 q ly1 m ,Ž .1 1 1

q 3q1 qq1 Ž .since t s l s 1. Then using 2.24 we get that1

r qq1 s mqly1 t q 2q1 q mqly1 t q 2yq q mqly1 tyqq1 q mly1 q s1 1 1

Ž . Ž . Ž .as required. Thus 2.22 and 2.24 imply 2.23 .

w q 2 yq xPROPOSITION 2.25. Let t s t , t , t g T be fixed. Also let u g U so1 1 1 1w x qq1 qthat u s r, s for some r and s such that r s s q s . Then the number of

u in U and n g T w such that un and t ha¨e the same characteristicl, m 0 l, m

equation is the following:Ž . Ž q 2 yq . Ž .2Ž .1 If t is of type t , t , t : q q 1 q y 1 .1 1 1

Ž . Ž . ŽŽ . Ž .Ž ..2 If t is of type t , t , t : q y 1 q q q y 1 q q 1 .1 1 1

Ž q 2 yq .Proof. If t is of type t , t , t then yl cannot be chosen to be an1 1 1eigenvalue since t / tyq. Thus there are q q 1 choices for l, q y 11 1choices for m, 1 choice for s, and q q 1 choices for r when t is of this type

Ž .by Proposition 2.19. Whereas if t is of type t , t , t we can choose1 1 1yl s t leaving q y 1 choices for m and 1 choice for each of r and s1since r must be zero by Corollary 2.17. Or we can choose yl / t of1which there are q choices and then there are q y 1 choices for m, 1 choicefor s, and q q 1 choices for r.

Now consider the maximal torus T . Recall that T is the maximal torus2 2w x qq1whose elements t are of the form t s t , t , t where t s 1. Thus1 2 3 i

JULIANNE G. RAINBOLT664

Ž . Ž . Ž .there are five types of elements in T : t , t , t , t , t , t , t , t , t ,2 1 2 3 1 1 3 1 2 1Ž . Ž . Ž .Ž .t , t , t , and t , t , t . Note that there are q q q 1 q y 1 elements t1 2 2 1 1 1

Ž . Ž .in T of type t , t , t , q q q 1 elements t in T in each of the types2 1 2 3 2Ž . Ž . Ž . Ž .t , t , t , t , t , t , t , t , t , and q q 1 elements t in T of type t , t , t .1 1 3 1 2 1 1 2 2 2 1 1 1

w xPROPOSITION 2.26. Fix an element t in T , so t s t , t , t for some2 1 2 3t , t , t g F 2

U with t qq1 s 1. Then un and t ha¨e the same characteristic1 2 3 q i l, m

equation if and only if the following two conditions occur:

s s t q t q t q l mq 2.27Ž . Ž .1 2 3

mqy1 s lty1 ty1 ty1 . 2.28Ž .1 2 3

Ž .Proof. Substituting t s a, t s b, and t s c immediately gives 2.271 2 3Ž . Ž . Ž . Ž . Ž .and 2.28 from 2.4 and 2.6 of Lemma 2.3. That is, 2.4 ] 2.6 of Lemma

2.3 become

s s t q t q t q l mq 2.29Ž . Ž .1 2 3

r qq1 s t t q t t q t t mq q m q sl ly1 2.30Ž . Ž .Ž .1 2 2 3 1 3

mqy1 s lty1 ty1 ty1 . 2.31Ž .1 2 3

Ž . Ž . Ž . Ž .Thus it remains to show that 2.29 and 2.31 imply 2.30 . Now 2.30requires that

r qq1 s mqly1 t t q t t q t t q mly1 q s.Ž .1 2 2 3 1 3

Ž .From 2.29 we have that

r qq1 s s q sq s s q t q q t q q t q q lq mq 2Ž .1 2 3

s s q ty1 q ty1 q ty1 q ly1 m ,Ž .1 2 3

q 2 U qq1 qq1 Ž .2since x s x for all x g F and since t s l s 1. Thus by 2.31 weq ihave that

qq1 q y1 q y1 q y1 y1r s m l t t q m l t t q m l t t q ml q s.1 2 2 3 1 3

w xPROPOSITION 2.32. Let t s t , t , t g T be fixed. Also let u g U so1 2 3 2w x qq1 qthat u s r, s for some r and s such that r s s q s . Then the number of

u in U and n g T w such that un and t ha¨e the same characteristicl, m 0 l, m

equation is the following:

Ž . Ž . Ž Ž . Ž .Ž .Ž ..1 If t is of type t , t , t : 3 q y 1 q q y 2 q y 1 q q 1 .1 2 3

Ž . Ž . Ž . Ž . Ž Ž . Ž2 If t is of type t , t , t , t , t , t , or t , t , t : 2 q y 1 q q y1 1 3 1 2 1 1 2 2.2Ž ..1 q q 1 .

Ž . Ž . ŽŽ . Ž .Ž ..3 If t is of type t , t , t : q y 1 q q q y 1 q q 1 .1 1 1

GELFAND]GRAEV REPRESENTATION 665

Proof. This proof is completely analogous to the proofs of Propositions2.18 and 2.25. Note that since each of t , t , and t are such that t qq1 s 1,1 2 3 iif yl is an eigenvalue of t it can be any of the t .i

2.7. The Homomorphism f : H ª CTT 00

Let u be an irreducible character of a maximal torus, T , of G. Then, asipreviously discussed, there exists a unique homomorphism f : H ª CT ,T ii

independent of u , which has the property that each homomorphismf : H ª C can be factored as f s u f . Also,T , u T Ti i , u i

f c s f c t t .Ž . Ž . Ž .ÝT Ti itgTi

One more lemma is needed before stating the results of the calculations ofŽ .Ž .the coefficients f c t . The proof of the following lemma is clear andTi

thus omitted.

LEMMA 2.33. Let u and n be fixed as in Subsections 2.5 and 2.6. Also letŽ y1 .t g T be fixed. Suppose there exists a g g G such that gung s t. Theni s

< ŽŽ . . < < Ž . < Ž y1 . y1C un s C t . Also hunh s t if and only if h s gx for someG s G sŽŽ . .x g C un .G s

Ž . Ž .In the remainder of this paper, the notation N x and T x will denotethe norm and trace of x g F 2 over F , respectively. Recall the basisq q

c , c N l, m g F 2U and lqq1 s 1� 4l, m l q

of the Hecke algebra described in Subsection 2.3. Here$

3 yqc s q e m , yl, m el, m

and

w xc s l, l, l e.l

Call c a basis element of the first type and c a basis element of thel, m l

second type.Ž .Ž .The next theorem gives the value of the coefficient f c t for i s 0, 1,T li

or 2 when c is a basis element of the second type.l

THEOREM 2.34. Let t be a fixed element of T , for i s 0, 1, or 2. Let ci l

be a fixed element of the basis of the Hecke algebra of the second type. Then

0, if t is not of type l, l, lŽ .f c t sŽ . Ž .T li ½ 1, if t is of type l, l, l .Ž .

JULIANNE G. RAINBOLT666

Proof. Within this proof c will be denoted by just c and T will bel iw xdenoted by just T. As before, let n s l, l, l , so that c s n e. Fix anl l

irreducible character, u of T. Let x be the unique irreducible character of² G : ² : Ž .G such that x , R / 0 and x , G / 0 see Subsection 1.2 . TheT , u

Ž . Ž . Ž .character table for G shows that x n s u n x I . Thusl l

< <y1 y1x n e s U c u x n uŽ . Ž . Ž .Ýl lugU

s u n x eŽ . Ž .l

s u n .Ž .l

Ž . Ž .Thus f c s u n . But also, by Curtis’s theorem,T , u l

f c s a u tŽ . Ž .ÝT , u ttgT

Ž Ž .for some coefficients a . That is, Curtis’s theorem writes f c ast T , u

Ž .Ž . Ž . .Ý f c t u t . Thust g T T

a u t s u n .Ž . Ž .Ý t ltgT

But this is true for all irreducible characters u of T. Thus a s 1 andnl

a s 0 for t / n .t l

w yq xTHEOREM 2.35. Again let t s t , t , t be a fixed element of T . But1 2 1 0now let c be a fixed element of the basis of the Hecke algebra of the firstl, m

type. Suppose that yl is an eigen¨alue of t. Then

q q 1 if t is of type yl, t , ylŽ .2f c t sŽ . Ž .T l , m0 ½ 1 if t is not of type yl, t , yl ,Ž .2

qy1 qy1 y1 Ž .Ž .if m s lt t . Otherwise, f c t s 0.1 2 T l, m0

Proof. In this proof T will be denoted by T. First suppose t is of type0Ž . Ž .t , t , t . Then 1.2 becomes1 1 1

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gungŽ . Ž . Ž . Ž .Ž .ÝT T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

3 y3 < <y1 y1 G y1s q ? q G c u Q gung .Ž . Ž .Ž .Ý T uggGugU

y1Ž .gung sts

GELFAND]GRAEV REPRESENTATION 667

Ž .But we also have by Lemma 2.16 that the 1, 2 entry of the matrix u mustŽ y1 .be zero in order for u to satisfy the condition gung s t since yl is ans

Ž y1 . Ž .eigenvalue of t. Thus c u s x 0 s 1 for every u in the above sum.Also by Corollary 2.10 the only u that meets the conditions of this sum isw Ž . q x Ž y1 .0, 3t q l m . For this u we must have that gung s t for all g g G,1 sby Lemma 2.33. Thus

< <y1 G y1f c t s G Q gung ,Ž . Ž . Ž .Ž .ÝT T uggG

w Ž . q x y1where u s 0, 3t q l m . First we will show that gung is not semisim-1Ž y1 . y1ple for any g g G such that gung s t. So suppose that gung iss

semisimple for some g. Then gungy1 s t. But t is in the center of G. Sothis implies un s t. But this is not possible. Thus gungy1 is not semisim-

Ž y1 . GŽŽ y1 . .ple. But this implies gung / I. So Q gung s 1. Thusu T u

< <y1 G y1f c t s G Q gung s 1.Ž . Ž . Ž .Ž .ÝT T uggG

Ž .This proves the theorem when t is of type t , t , t .1 1 1Ž yq .Now suppose t is of type t , t , t . Then1 2 1

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gungŽ . Ž . Ž . Ž .Ž .ÝT T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

3 y3 < <y1 y1s q ? q T c uŽ .ÝggG

y1Ž .gung sts

Ž .since now C t s T by Subsection 2.4, u is completely determinedGT Ž .by the same argument as above, and Q x s 1 for all x. But, as aboveT

Ž y1 . Ž .c u s x 0 s 1. Thus

< <y1f c t s T 1.Ž . Ž . ÝTggG

y1Ž .gung sts

This in turn implies

< <y1f c t s T 1 s 1,Ž . Ž . ÝTŽŽ . .xgC unG s

Ž yq .by Lemma 2.33. This proves the theorem when t is of type t , t , t .1 2 1

JULIANNE G. RAINBOLT668

Ž . Ž y1 .Now suppose t is of type t , t , t . We still have that c u s 1 but1 2 1Ž . Ž .3Ž .now C t s H = S which has order q q q 1 q y 1 by SubsectionG 2 2

2.4. Thus, in this case,

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gungŽ . Ž . Ž . Ž .Ž .ÝT T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

y13 H y12s q q q 1 q y 1 Q gung .Ž . Ž . Ž .Ž .Ž . Ý T uggGugU

y1Ž .gung sts

Ž Ž y1 . Ž . w xHere gung g C t so it is a matrix of the form 0, x , by Subsectionu G.2.4. Now

Q H 2 gungy1 s q q 1Ž .Ž .T u

if and only if x s 0. But this is true if and only if gungy1 is semisimple.Suppose that gungy1 is semisimple for some g in the above sum. Thengungy1 s t and thus gun s tg. That is,

yq yq at bt ctasm q cm ybl am 1 1 1yq yq dt et ftdsm q fm yel dm s ,2 2 2yq yq� 0 � 0hsm q jm yil hm ht it jt1 1 1

a b cd e fwhere g s .� 0h i j

But yl is an eigenvalue, so first suppose that yl s t and thus yl / t ,2 1Ž .since t is of type t , t , t . A comparison of the second columns of these1 2 1

two matrices gives b s i s 0. Then by comparing the first and thirdcolumns we get that both matrices must be equal to

at 0 am1

dt et dm .2 2� 0ht 0 hm1

But this matrix has determinant zero, a contradiction. Thus yl / t . Thus2if gungy1 is semisimple then yl s t and thus yl / t . By Subsection1 2

Ž . q Ž . q2.5, s s 2 t q t q l m s t q t m . Now let1 2 1 2

1 0 ty1m1

y1g s .1 0 t m2� 00 1 0

GELFAND]GRAEV REPRESENTATION 669

Ž . y1Note that det g is not zero since t / t . Now we will show that gung s t.2 1Note that

t q t q ty1myqq1 0 m1 2 1

y1 yqq1gun s t q t q t m 0 m1 2 2� 00 yl 0

t q t q ty1 yt t 0 mŽ .1 2 1 1 2

y1s ,t q t q t yt t 0 mŽ .1 2 2 1 2� 00 t 01

since

myqq1 s ly1 t 2 t s yty1 t 2 t s yt t1 2 1 1 2 1 2

by Subsection 2.5 and since yl s t . Thus1

t 0 m1

t 0 mgun s .2� 00 t 01

Also

t 0 m1

t 0 mtg s .2� 00 t 01

Thus gun s tg as claimed. Thus when yl s t there exists a g such that1y1 Ž . y1 Ž y1 . y1gung is semisimple. But then g un g s gung s gung . Thuss s

Ž . H 2Ž y1 .un s un. So un is semisimple. This implies Q gung s q q 1 for alls TŽ . H 2Ž y1 .g in the above sum if t is of type yl, t , yl , otherwise Q gung s 1.2 T

Thus

y13 H y12f c t s q q q 1 q y 1 Q gungŽ . Ž . Ž . Ž . Ž .Ž .Ž . ÝT T uggG

y1Ž .gung sts

y13s q q q 1 q y 1 1Ž . Ž .Ž . ÝggG

y1Ž .gung sts

JULIANNE G. RAINBOLT670

Ž . Ž .if t is of type t , t , t but not of type yl, t , yl and1 2 1 2

y13f c t s q q q 1 q y 1 q q 1Ž . Ž . Ž . Ž . Ž .Ž . ÝTggG

y1Ž .gung sts

Ž .if t is of type yl, t , yl . But by Lemma 2.33 these sums are over a set2< ŽŽ . . < Ž .3Ž . Ž .Ž .of size C un s q q q 1 q y 1 . Thus f c t s 1 if t is of typeG s T

Ž . Ž . Ž .Ž .t ,t , t but not of type yl, t , yl and f c t s q q 1 if t is of type1 2 1 2 TŽ .yl, t , yl .2

We will now prove a similar result where we instead assume that the ylentry of n s n is not an eigenvalue of the fixed t g T .l, m 0

w yq xTHEOREM 2.36. Let t s t , t , t be a fixed element of T . Let c be1 2 1 0 l, m

a fixed element of the basis of the Hecke algebra of the first type. Suppose thatŽ yq . qyl is not an eigen¨alue of t. Let s s t q t q t q l m . Then1 2 1

y1w xf c t s c r , s s x yr ,Ž . Ž . Ž .Ž .Ý ÝT l , m2 2rgF rgFq q

Ž . Ž . Ž . Ž .N r sT s N r sT s

qy1 qy1 y1 Ž .Ž .if m s lt t . Otherwise, f c t s 0.1 2 T l, m

Ž .Proof. Again consider 1.2 ,

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gung .Ž . Ž . Ž . Ž .Ž .ÝT T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

The proof of the previous theorem almost works here again except for theplaces where we used that yl is an eigenvalue. This was used in twodifferent places. One was in determining the values of the Green functions

Ž y1 . Ž y1 .on gung and the other was to conclude that c u was 1. FirstuŽ yq . CGŽ t .Ž .consider the first of these two. If t is of type t , t , t then Q x s1 2 1 T0T0Ž . Ž yq .Q x s 1 for all x. Thus when t is of type t , t , t the Green functionT 1 2 10

Ž . Ž .in 1.2 will always take the value 1. So suppose that t is of type t , t , t1 2 1Ž . Ž . y1or t , t , t and that g is included in 1.2 with gung semisimple. Then1 1 1

similar to the above argument we must have gun s tg. That is,

yq q yq yq at bt ctasm q br m q cm yarl y bl am 1 1 1yq q yq yq dt et ftdsm q er m q fm ydrl y el dm s ,2 2 2yq q yq yq� 0 � 0hsm q ir m q jm yhrl y il hm ht it jt1 1 1

GELFAND]GRAEV REPRESENTATION 671

where

a b cd e fg s � 0h i j

and we allow t s t . A comparison of the last columns of these two1 2matrices shows that

j s hm ty1 and c s am ty1 .1 1

A comparison of the second columns shows that

yhrl s it q il s i t q l and yarl s bt q bl s b t q l .Ž . Ž .1 1 1 1

Ž .y1But yl is not an eigenvalue of t so t q l / 0. Thus i s yhrl t q l1 1Ž .y1and b s yarl t q l . Substituting these expressions for b, c, i, and j1

into the matrix on the right hand side we get

y1at yart l t q l amŽ .1 1 1

dt et ft .2 2 2

y1� 0ht yhrt l t q l hmŽ .1 1 1

But this matrix is singular since the first and third rows are multiples ofeach other. This is a contradiction. So gungy1 is not semisimple. Thus the

H 2ŽŽ y1 . . GŽŽ y1 . .values of the Green functions Q gung , Q gung , andT u T uT ŽŽ y1 . .Q gung will always be 1.T uNow for the other adjustment that has to be made to the previous proof.

y1 Ž .By Corollary 2.10 in order for gung to be included in 1.2 we must haveŽ yq . qthat the entry s s t q t q t q l m . This corollary also showed that1 2 1

there are no restrictions on r except for the condition that r qq1 s s q sq

Ž Ž . Ž ..i.e., N r s T s which just comes from the requirement that u g U.Thus

ind nŽ . y1 C Ž t . y1Gf c t s c u Q gungŽ . Ž . Ž . Ž .Ž .ÝT T uG² : < < < <Q , G U C tŽ .T G ggGugU

y1Ž .gung sts

3 y3 < <y1 y1s q ? 1 ? q C t c uŽ . Ž .ÝGŽŽ . .xgC unG s

ugUŽ . Ž .N r sT s

y1s c u .Ž .ÝugU

Ž . Ž .N r sT s

JULIANNE G. RAINBOLT672

Ž .Ž .Note. In summary the coefficients f c t are equal to:T0

Ž .If c is of the first type c s c ,l, m

¡q q 1, if t s yl, t / yl, and N n s N tŽ . Ž .1 2 l , m

1, if t s yl and N n s N tŽ . Ž .2 l , m

~ x yr , if t / yl and N n s N tŽ . Ž . Ž .Ý i l , m2rgFq

Ž . Ž .N r sT s¢0, if N n / N t .Ž . Ž .l, m

Ž .If c is of the second type c s c ,l

0, if t is not of type l, l, lŽ .½ 1, if t is of type l, l, l .Ž .

2.8. The Homomorphism f : H ª CTT 11

Ž .Ž .Now we will compute the coefficients f c t when i s 1 by proving theTi

analogues of the theorems in Subsection 2.7 when T s T . Note thati 1Lemma 2.33 and Theorem 2.34 did not assume a particular maximal torus.Thus we will begin with the theorem parallel to Theorem 2.35.

w q 2 yq xTHEOREM 2.37. Let t s t , t , t be a fixed element of T . Let c be1 1 1 1 l, m

a fixed element of the basis of the Hecke algebra of the first type. Suppose thatŽ .Ž . Ž . qy1 yq 2qq y1yl is an eigen¨alue of t. Then f c t s y q q 1 if m s lt .T l, m 11

Ž .Ž .Otherwise, f c t s 0.T l, m1

Proof. Notice that since yl is an eigenvalue we must have that t is ofŽ . qq1type yl, yl, yl . This follows from the assumptions that l s 1 and

t q 3q1 s 1. For if t s yl then t qq1 s 1 and so t s tyq. But, as men-1 1 1 1 1tioned in Subsection 2.4, if any two of the diagonal elements of t are equalthen all three are equal. Thus assuming t s yl forces t to be of type1

Ž . q 2 q 3qq 2yl, yl, yl . Similarly, if we assume t s yl, then t s 1. Thus1 1y1qq 2 q 2 Ž .t s 1 and thus t s t , again forcing t to be of type yl, yl, yl .1 1 1

If we assume tyq s yl, then tyq 2yq s 1. Thus tyq s t q 2. So we again get1 1 1 1

Ž .that t is of type yl, yl, yl . Now, as in the first part of the proof ofŽ ² G : . Ž .Theorem 2.35 with the adjustment Q , G s y1 instead of 1 , 1.2T1

becomes

< <y1 G y1f c t s y G Q gung ,Ž . Ž . Ž .Ž .ÝT T u1 1ggG

GELFAND]GRAEV REPRESENTATION 673

w Ž . q xwhere u s 0, y2l m . Also, as the proof of Theorem 2.35 shows,y1 ˜Ž .gung is not semisimple. Note that un is conjugate in G to

yl 1 0x s .0 yl 0ž /0 0 yl

In fact, let

yly1 0 0y11 0 ymlg s .� 0y11 1 yml

But mqy1 s lty3 s yly2 . Thus1

2 0 ymly1

gun s .yl 0 m� 0yl yl m

Also

2 0 ymly1

xg s .yl 0 m� 0yl yl m

y1 GŽŽ y1 . .Thus gung s x. Thus Q gung s q q 1. SoT u1

< <y1 G y1f c t s y G Q gungŽ . Ž . Ž .Ž .ÝT T u1 1ggG

y1< <s y G q q 1 s y q q 1 .Ž . Ž .ÝggG

w q 2 yq xTHEOREM 2.38. Again let t s t , t , t be a fixed element of T . Also1 1 1 1let c be a fixed element of the basis of the Hecke algebra of the first type.l, m

Ž q 2But now suppose that yl is not an eigen¨alue of t. Let s s t q t q1 1yq . qt q l m . Then1

y1w xf c t s y c r , s s y x yr .Ž . Ž . Ž .Ž .Ý ÝT l , m2 2rgF rgFq q

Ž . Ž . Ž . Ž .N r sT s N r sT s

Proof. Now the proof of Theorem 2.36 almost translates to a proof ofthis theorem by just changing T to T except that in this case the Green0 1functions are not determined to be 1 just by knowing that gungy1 is not

JULIANNE G. RAINBOLT674

Ž . Ž Ž q 2 yq ..semisimple. In the case C t s T i.e., t is of type t , t , t we stillG 1 1 1 1have that QCGŽ t . s QT1 which is always 1. But suppose that t is of typeT T1 1

Ž .t , t , t . In this case it is necessary to show that un is also not conjugate1 1 1˜Ž .in G to

t 1 01

0 t 01� 00 0 t1

in order to conclude that the Green function is always 1. So suppose un isconjugate to

t 1 01

0 t 0x s 1� 00 0 t1

and let

a b cd e fg s � 0h i j

be the matrix such that gun s xg. This equation in matrices is

asmyq q br qmyq q cmyq yarl y bl amyq q yq yqdsm q er m q fm ydrl y el dmyq q yq yq� 0hsm q ir m q jm yhrl y il hm

at q d bt q e ct q f1 1 1

dt et fts .1 1 1� 0ht it jt1 1 1

A comparison of the last columns of these two matrices shows thatj s hm ty1 and f s dm ty1. A comparison of the second columns shows1 1

Ž . Ž .that ydrl s et q el s e t q l and hrl s it q il s i t q l . But yl1 1 1 1Ž .y1is not an eigenvalue of t so t q l / 0. So i s yhrl t q l and1 1

Ž .y1e s ydrl t q l . Substituting these expressions for j, f , i, and e into1the matrix on the right hand side gives

at q d bt q e ct q f1 1 1

y1dt ydrt l t q l dmŽ . .1 1 1

y1� 0ht yhrt l t q l hmŽ .1 1 1

GELFAND]GRAEV REPRESENTATION 675

But the second and third rows of this matrix are multiples of each other, soit is not invertible. This is a contradiction. So un is not conjugate to x.

Ž .Thus as in Theorem 2.36, the values of the Green functions in 1.2 willalways be 1. The rest of the proof of Theorem 2.36 goes through hereverbatim.

Ž .Ž .Note. In summary the coefficients f c t are equal to:T1

Ž .If c is of the first type c s c ,l, m

¡y q q 1 , if yl an eigenvalue and N n s N tŽ . Ž . Ž .l, m

y x yr , if yl not an eigenvalue and N n s N tŽ . Ž . Ž .Ý l, m~2rgFq

Ž . Ž .N r sT s¢0, if N n / N t .Ž . Ž .l, m

Ž .If c is of the second type c s c ,l

0, if t is not of type l, l, lŽ .½ 1, if t is of type l, l, l .Ž .

2.9. The Homomorphism f : H ª CTT 22

Ž .Ž .Now we will compute the coefficients f c t when i s 2. TheoremTi

2.34 computed these coefficients when c is a basis element of the secondtype. The following two theorems compute these coefficients when c is ofthe first type.

w xTHEOREM 2.39. Let t s t , t , t be a fixed element of T . Let c be a1 2 3 2 l, m

fixed element of the basis of the Hecke algebra of the first type. Suppose thatyl is an eigen¨alue of t. Then

2 q y 1 if the eigen¨alue yl occurs with multiplicity 3¡~y qq1 if the eigen¨alue yl occurs with multiplicity 2f c t s Ž .Ž . Ž .T l , m2 ¢y1 if the eigen¨alue yl occurs with multiplicity 1,

qy1 y1 y1 y1 Ž .Ž .if m s lt t t . Otherwise, f c t s 0.1 2 3 T l, m2

Ž .Proof. First suppose t is of type t , t , t . Then, as in the first part of1 1 1Ž ² G :the proof of Theorem 2.35 with the adjustment Q , G s y1 instead ofT2

. Ž .1 , 1.2 becomes

< <y1 G y1f c t s y G Q gung ,Ž . Ž . Ž .Ž .ÝT T u2 2ggG

JULIANNE G. RAINBOLT676

w Ž . q xwhere u s 0, 3t q l m . Also, as the proof of Theorem 2.35 shows,1gungy1 is not semisimple. As shown in the proof of Theorem 2.37, un isconjugate to

t 1 01

0 t 0 .1� 00 0 t1

GŽŽ y1 . .Thus Q gung s y2 q q 1. So thatT u2

< <y1 G y1f c t s y G Q gungŽ . Ž . Ž .Ž .ÝT T u2 2ggG

< <y1s y G y2 q q 1 s 2 q y 1.Ž .ÝggG

Ž .This proves the theorem when t is of type t , t , t .1 1 1Ž .Now suppose t is of type t , t , t . For t of this type the proof of1 2 3

Theorem 2.35 translates to prove this theorem by just changing everywhereyou see T to T and adjusting by a multiple of y1.0 2

Ž .Similarly the proof when t is of type t , t , t is exactly the same as the1 2 1proof of Theorem 2.35 where we showed that gungy1 is semisimple. ThusQ H 2 s q q 1. Also by symmetry that proof translates into a proof of thisT2

Ž . Ž .theorem when t is of type t , t , t or of type t , t , t .1 1 3 1 2 2

w xTHEOREM 2.40. Again let t s t , t , t be a fixed element of T . Also let1 2 3 2c be a fixed element of the basis of the Hecke algebra of the first type. Butl, m

Ž . qnow suppose that yl is not an eigen¨alue of t. Let s s t q t q t q l m .1 2 3Then

y1w xf c t s y c r , s s y x yr ,Ž . Ž . Ž .Ž .Ý ÝT l , m2 2rgF rgFq q

Ž . Ž . Ž . Ž .N r sT s N r sT s

qy1 y1 y1 y1 Ž .Ž .if m s lt t t . Otherwise, f c t s 0.1 2 3 T l, m

Proof. Now the proofs of Theorems 2.36 and 2.38 almost translate to aproof of this theorem by just changing T or T to T . In the case0 1 2

Ž . Ž Ž .. CGŽ t . T2C t s T i.e., t is of type t , t , t we still have that Q s Q s 1.G 2 1 2 3 T T1 2

Ž . Ž . Ž .In the case t is of type t , t , t , t , t , t , or t , t , t it is enough to1 2 1 1 1 3 1 2 2

show gungy1 is not semisimple to conclude QCGŽ t . s Q H i s 1 as was doneT T2 2

in Theorem 2.36. In Theorem 2.38 it was shown that if t is of typeŽ . Ž .t , t , t , the value of the Green functions in 1.2 will still always be 1.1 1 1

Ž .Thus as in Theorem 2.36, the values of the Green functions in 1.2 willalways be 1. The rest of the proof of Theorem 2.36 goes through hereverbatim.

GELFAND]GRAEV REPRESENTATION 677

Ž .Ž .Note. In summary the coefficients f c t are equal to:T2

Ž .If c is of the first type c s c ,l, m

Ž . Ž .2 q y 1, if yl a multiplicity 3 eigenvalue and N n sN t¡ l, m

Ž . Ž . Ž .y q q 1 , if yl a multiplicity 2 eigenvalue and N n sN tl, m

Ž . Ž .y1, if yl a multiplicity 1 eigenvalue and N n sN tl, m~Ž . Ž . Ž .y x yr , if t / yl and N n sN tÝ i l , m

2rgFqŽ . Ž .N r sT s¢ Ž . Ž .0, if N n / N t .l, m

Ž .If c is of the second type c s c ,l

0, if t is not of type l, l, lŽ .½ 1, if t is of type l, l, l .Ž .

The results of Subsections 2.7, 2.8, and 2.9 give us the homomorphismsf : H ª CT . Composing these homomorphisms with the irreducible char-T ii

acters of T will give all the irreducible characters of H, by Curtis’sitheorem, Theorem 1.1.

3. THE STRUCTURE CONSTANTS OF THE HECKEALGEBRA H

In this section we will continue to use the notation given in Section 2. Asexplained in Subsection 2.3, if

$3 yqc s q e t , yu, tu , t

andw xc s u , u , u eu

then

c , c N t , u g F 2U and uqq1 s 1� 4u , t u q

is a basis for H.Ž .To simplify notation, let J be an index set of cardinality q q 1 q

Ž 2 .Ž . Ž .Ž 2 .q y 1 q q 1 s q q 1 q . For j F q q 1 let

w xx s u , u , u ,j j j j

U Ž .2where u runs over the elements of F such N u s 1. For j ) q q 1 letj q j$

yqx s t , yu , t ,j j j j

JULIANNE G. RAINBOLT678

U Ž .2where again u runs over the elements of F such N u s 1 and t runsj q j jU Ž .2over the elements of F . Note that ind x s 1 for j less than or equal toq j

Ž . 3 Ž .q q 1 and ind x s q for j greater than q q 1. Now let a s ind x ex e.j j j j� 4Then a is the same basis of H as described above.j jg J

Now given two basis elements a and a of the Hecke algebra H, theiri jproduct

a a s m aÝi j i jk kkgJ

for some structure constants m . The purpose of this section is toi jkexplicitly compute these constants.

Ž . Ž .In the three cases, 1 i F q q 1, j F q q 1, 2 i F q q 1, j ) q q 1,Ž .and 3 i ) q q 1, j F q q 1, this computation is trivial. First note that the

Ž . Ž .Hecke algebra is commutative so that cases 2 and 3 are the same.Ž .Clearly in case 1 , a a s a where u u s u . That is, c c s c . Ini j k i j k u u u ui j i j

Ž .case 2 we have that$

yqa a s a , where x s u t , yu u , u t .i j k k i j i j i j

That is, c c s c . Thus the only interesting case is when both iu u , t u u , u ti j j i j i j

and j are greater than q q 1. So in the remainder of this section assumethat i and j are both greater than q q 1.

The elements of the group algebra CG can be identified with the set offunctions f : G ª C, where Ý a x g CG corresponds to the functionx g G x

Ž .f : G ª C defined by f x s a . Using this correspondence, the structurexconstants

< < y1m s U a y a y x ,Ž . Ž .Ýi jk i j ky1ygD lx Di k j

where D is the double coset Ux U and Dy1 is the double coset Uxy1Ui i j jŽ w x.see, for example, 4, p. 280 .

First we will compute m when k F q q 1. To do this we need toi jkdetermine when y g D l x Dy1. Now since i ) q q 1 we have thati k j$

yqw xx s t , yu , t . Thus D s Ux U si i i i i i

¡ yq yq yq qbt bct y au bdt y ac u q ti i i i i iq yq q yq q yq q~ a t a ct y u a dt y c ui i i i iyq yq yq¢� 0t ct dti i i

¦¥N N a s T b , N c s T d .Ž . Ž . Ž . Ž .§

GELFAND]GRAEV REPRESENTATION 679

Also since k F q q 1 and j ) q q 1 we have that

w xx s u , u , uk k k k

and$

y1 q y1 y1x s t , yu , t .j j j j

Thus x Dy1 equalsk j

¡ y1 y1 y1 q q y1 y1yu t yxu u q yzu t u t y xz u u q ywu tk j k j k j k j k j k j

q y1 y1 q y1 q y1 q y1~ x u t yu u q x zu t yz u u q x wu tk j k j k j k j k j

y1 y1 y1� 0¢ u t zu t wu tk j k j k j

¦¥N N x s T y , N z s T w .Ž . Ž . Ž . Ž .§

Ž . yq y1A comparison of the 3,1 entries reveals that t s u t for an elementi k jto be in both D and x Dy1. Then using the fact that none of t , t , u arei k j i j kzero we get from comparison of the first columns and third rows thatb s y, aq s x q, c s z, and d s w. But we already have that aqq1 s b q bq.Thus aqq1 s y q y q s x qq1 s aq x. Thus a s 0 or a s x. But if a s 0then aq s 0 so that x q s 0 and thus x s 0. So a s x. Thus if u x u s1 i 2x u xy1 u for some u g U then u s u and u s u . But x is central.k 3 j 4 i 1 3 2 4 kThus x s x xy1. That is,i k j

$$yq q y1 y1t , yu , t s u t , yu u , u t .i i i k j k j k j

Thus u s u u s t tyq s tyq t . But u u s tyq t implies u u t q s t . Sok i j i j i j i j i j i j i j

that uy1 uy1 t s t q since uq s uy1, uq s uy1 and t q 2 s t . This impliesi j i j i i j j i it tyq s u u . Thus u u s tyq t implies u u s t tyq. Similarly u u s t tyqi j i j i j i j i j i j i j i j

implies u u s tyq t . Thus D l x Dy1 s B unless u s u u s tyq t . Thati j i j i k j k i j i jy1 w xis, D l x D s B unless x s u u , u u , u u .i k j k i j i j i j

So from now on assume u s u u s tyq t . Then by directly multiplyingk i j i jthe matrices we see that D s x Dy1. Thus D l x Dy1 s D . So m si k j i k j i i jk< < Ž . Ž y1 .U Ý a y a y x .y g D i j ki

Now

3 3 < <y1 y1 < <y1 y1a s q ex e s q U c u ux U c ¨ ¨Ž . Ž .Ý Ýi i i¨gU ¨gU

s qy3 c uy1 ¨y1 ux ¨ .Ž .Ý iu , ¨

JULIANNE G. RAINBOLT680

Ž .Also if y g D then y s u x ¨ for some u , ¨ g U. Thus a y si y i y y y iy3 Ž y1 y1. y1 Ž .y1 y1 y1 y1q c u ¨ . Also notice that y x s u x ¨ x s ¨ x u xy y k y i y k y i y k

s ¨y1 xy1 x uy1 s ¨y1 x uy1, since x is central and x s xy1 x byy i k y y j y k j i k

Ž y 1 . y 3 Ž . Ž . Ž y 1 .above. Thus a y x s q c ¨ u . Thus a y a y x sj k y y i j ky3 Ž y1 y1. y3 Ž . y6 3 y6 3q c u ¨ q c ¨ u s q . Thus m s q Ý q s q when uy y y y i jk y g D ki

s u u s tyq t , and m s 0 otherwise. Leti j i j i jk

0 if r / sd r , s sŽ . ½ 1 if r s s.

Ž yq . 3Then m s d u u , t t q only for the one index number k F q q 1i jk i j i jsuch that u s u u . Otherwise m s 0 when k F q q 1.k i j i jk

It remains to consider m when i, j, and k are all greater than q q 1.i jkIn this case we have

$yqD s Ux U s U t , yu , t U,i i i i i

$y1 y1 q y1 y1D s Ux U s U t , yu , t Uj j j j j

and

$yqx s t , yu , t .k k k k

Suppose that y g D l x Dy1. Then y s wx z s x uxy1 ¨ for somei k j i k j

u, ¨ , w, z g U. So y¨y1 s wx zX s x uxy1 for some u, ¨ , w, zX g U. Thusi k j

$$ $yq yq q y1 y1w x w x w xa, b t , yu , t c, d s t , yu , t r , s t , yu , t ,i i i k k k j j j

Ž . Ž . Ž . Ž .for some a, b, c, d, r, s g U such that N a s T b , N c s T d , andŽ . Ž .N r s T s . Then

btyq bctyq y au bdtyq y acqu q ti i i i i iq yq q yq q yq qa t a ct y u a dt y c ui i i i iyq yq yq� 0t ct dti i i

ty1 t 0 0j k

q y1 y1yr t u u u 0s .j k j k

y1 yq y1 yq q yq� 0st t yru t t tj k j k j k

GELFAND]GRAEV REPRESENTATION 681

Thus s s tyq t t q, b s t qty1 t , and d s t qt qtyq. Also bctyq y au s 0i j k i j k i j k i i

so that au s bctyq s ct qty1 t tyq s cty1 t . Thus a s cty1 t uy1. Also r si i i j k i j k j k i

yctyq t qu . Substituting these values for a, b, d, r, and s in the abovei k jmatrix equation gives the new equation

ty1 t 0 t qt qy1 tyqq1 y cqq1 ty1 t q tj k i j k j k i

q yq yq q qq1 yq yq qc t t t u c t t t u y u 0i j k i i j k i i

yq yq q yq� 0t ct t ti i j k

ty1 t 0 0j k

q y1 y1 y1 y1c t t t u u u u 0s .i j k j k j k

yq yq q yq� 0t ct t ti i j k

But cqq1 s d q dq s t qt qtyq q t t ty1. Thus t qt qy1 tyqq1 y cqq1 ty1 t q ti j k i j k i j k j k i

s t qt qy1 tyqq1 y t qt qy1 tyqq1 y t q t s 0 and cqq1 tyq tyq t qu y u s ui j k i j k i i i j k i i i

qtyqq1 tyqq1 t qy1u y u s tyqq1 tyqq1 t qy1u . Soi j k i i i j k i

ty1 t 0 0j k

q yq yq q yqq1 yqq1 qy1c t t t u t t t u 0i j k i i j k i

yq yq q yq� 0t ct t ti i j k

ty1 t 0 0j k

q y1 y1 y1 y1c t t t u u u u 0s .i j k j k j k

yq yq q yq� 0t ct t ti i j k

yqq1 yqq1 qy1 Ž .Thus u s t t t u u . Notice that now the 2,1 entries can bothk i j k i j

be simplified to cqtyq tyq t qu by substituting in this expression for u . Alsoi j k i kŽ . Ž . Ž . Ž . Ž . Ž .note that the requirements N a s T b , N c s T d , and N r s T s

are satisfied when we take the above formulas for a, b, c, d, r, and s.Thus the above shows that m will be nonzero only when x is of thei jk k$

yqq1 y qq1 qy1 yqw xform t , yt t t u u , t , i.e., when the u entry of x isk i j k i j k k k

tyqq1 tyqq1 t qy1u u . So we fix an x of this form.i j k i j kLet

y1 y1 q y1 w xA s ct t u , t t t s a, b ,c j k i i j k

q q yq w xB s c, t t t s c, d ,c i j k

JULIANNE G. RAINBOLT682

andyq q yq q w xC s yct t u , t t t s r , s ,c i k j i j k

where c g F 2 is such that cqq1 s t qt qtyq q t t ty1.q i j k i j kThen the above calculations show that

Ux U l x Uxy1UŽ . Ž .i k j

s y N y¨y1 g Ux U l x Uxy1 , ¨ g UŽ .½ 5Ž .i k j

s y N y¨y1 g A x B , ¨ g U, cqq1 s t qt qtyq q t t ty1� 4c i c i j k i j k

s y N y¨y1 g x C xy1 , ¨ g U, cqq1 s t qt qtyq q t t ty1 .� 4k c j i j k i j k

So suppose that y is in the above set. So y s A x B ¨ s x C xy1 ¨c i c k c j

for some ¨ g U and for some c such that cqq1 s t qt qtyq q t t ty1. Now,i j k i j kusing the notation explained earlier,

< <y1 y1 y1a s U c u w ux w.Ž .Ýi iu , wgU

Ž . < <y1 Ž y1 . ŽŽ .y1 .Thus a y s U c A c B ¨ . Alsoi c c

< <y1 y1 y1a s U c u w ux wŽ .Ýj ju , wgU

and y s x C xy1 ¨ . So that yy1 x s ¨y1 x Cy1. Thusk c j k j c

y1 y1 y1 < <y1a y x s a ¨ x C s U c ¨ c C .Ž . Ž .Ž . Ž .j k j j c c

Thus

< < y1m s U a y a y xŽ . Ž .Ýi jk i j ky1D lx Di k j

< < < <y1 y1 y1s U U c A c ¨Ž .Ž .Ý Ý cc¨gU

qq1 q q yq y1c st t t qt t ti j k i j k

= y1 < <y1c B U c ¨ c CŽ . Ž .Ž .c c

< <y1 y1 y1s U c A B CŽ .Ý Ý c c cc¨gU

s c Ay1By1 C .Ž .Ý c c cc

qq1 q q yq y1c st t t qt t ti j k i j k

GELFAND]GRAEV REPRESENTATION 683

y1 y1 w Ž y1 y1 yq q . xBut A B C s yc 1 q t t u q t t u , ) , where the ) entryc c c j k i i k jis an expression involving t , t , u , u , t , and c. Recall there exists ani j i j k

Žw x. Ž .2additive character x of F such that c a, b s x a . Thusq

m s c Ay1By1 CŽ .Ýi jk c c cc

qq1 q q yq y1c st t t qt t ti j k i j k

s x yc 1 q ty1 t uy1 q tyq t quŽ .Ý ž /j k i i k jc

qq1 q q yq y1c st t t qt t ti j k i j k

s x ya x 1 q ty1 t uy1 q tyq t qu ,Ž .Ý ž /k j k i i k jŽ .N x s1

where in the last summation a is a fixed element of F 2 such thatk qaqq1 s t qt qtyq q t t ty1.k i j k i j k

Combining all of the above calculations gives the following proposition,which describes the structure constants.

� 4PROPOSITION 3.1. Let c , c be the basis for the Hecke Algebra de-u, t uscribed abo¨e. Then

c c s c , 3.2Ž .u ¨ u¨

c c s c , 3.3Ž .u , t ¨ ¨ u , ¨ t

c c s q3d u u , tyq t cŽ .u , t u , t 1 2 1 2 u u1 1 2 2 1 2

q x ya x 1 q ty1 t uy1 q tyq t qu c ,Ž .Ž .Ý Ý k 2 k 1 1 k 2 u , tk kU Ž .N x s12t gFk q

3.4Ž .

Ž . yqq1 yqq1 qy1where in 3.4 , u s t t t u u and a is fixed for a fixed t and isk 1 2 k 1 2 k ksuch that aqq1 s t qt qtyq q t t ty1.k 1 2 k 1 2 k

Ž .For a possibly more usable form of 3.4 of this proposition make thechange of variable r s ty1 ty1 t in the equation1 2 k

c c s q3d u u , tyq t c qŽ . Ýu , t u , t 1 2 1 2 u u1 1 2 2 1 2U

2t gFk q

= x yc 1 q ty1 t uy1 q tyq t qu c .Ž .Ž .Ý 2 k 1 1 k 2 u , tk kc

qq1 q q yq y1c st t t qt t t1 2 k 1 2 k

JULIANNE G. RAINBOLT684

qy1 Ž .Then u s r u u and 3.4 becomesk 1 2

c c s q3d u u , tyq t c qŽ . Ýu , t u , t 1 2 1 2 u u1 1 2 2 1 2U

2rgFq

= x yc 1 q rt uy1 q r qt qu c qy 1 .Ž .Ž .Ý 1 1 2 2 r u u , r t t1 2 1 22cgFq

y1Ž . Ž .N c sT r

Ž .For another form of 3.4 we can rewrite the preceding equation interms of matrices. Let

yqK s t , u , t ,i i i i

w yq xR s r , 1, r ,

and

˜ qy1 yqw xR s r , r , r .

� Ž . 4Also let U s A g U N the 1,3 entry of A is r .rŽ . Ž .y1 Ž .q y1Ž .yq Ž .Then A s K R B K R and C s K R B K R . Thus 3.4c 1 c 1 c 2 c 2

becomes

c c s c c s q3d u u , tyq t cŽ .u , t u , t K K 1 2 1 2 u u1 1 2 2 1 2 1 2

y1y1 y1q c K R B K R c BŽ . Ž . Ž .Ž .Ý Ý 1 1U y1BgU2rgF rq

=q yqy1c K R B K R c .Ž . Ž .Ž . ˜2 2 R K K1 2

Ž .Thus 3.4 becomes

c c s q3d u u , tyq t c q c K1 R By1Ž .Ž . Ý ÝK K 1 2 1 2 u u1 2 1 2U y1BgU2rgF rq

=c By1 c ŽK 2 R.qBy1 c .Ž . Ž . R K K1 2

REFERENCES

1. R. W. Carter, ‘‘Finite Groups of Lie Type: Conjugacy Classes and Complex Characters,’’Wiley]Interscience, New York, 1985.

Ž . Ž .2. R. W. Carter, Cuspidal matrix representations for GL q and GL q , Proc. London2 3Ž . Ž .Math. Soc. 3 64 1992 , 487]523.

3. C. W. Curtis, On the Gelfand]Graev representations of a reductive group over a finiteŽ .field, J. Algebra 157 1993 , 517]533.

GELFAND]GRAEV REPRESENTATION 685

4. C. W. Curtis and I. Reiner, ‘‘Methods of Representation Theory,’’ Vol. 1, Wiley]Intersci-ence, New York, 1981.

5. F. Digne and J. Michel, ‘‘Representations of Finite Groups of Lie Type,’’ LondonMathematical Society Student Texts, Vol. 21, Cambridge Univ. Press, LondonrNew York,1991.

6. V. Ennola, On the characters of the finite unitary groups, Ann. Acad. Sci. Fenn. Ser. AŽ .323 1963 , 1]35.

7. R. Steinberg, Lectures on Chevalley groups, mimeographed notes, Yale University, 1967.Ž 2 .8. D. Surowski, On a class of principal series characters of SU 3, q , preprint, Kansas State

University, 1976.